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I came upon this while doing waves:

How do you write $x \sin A+y \sin B$ in the form of $z\sin(\cdots)$?

I have absolutely no idea on how to proceed.

Edit: It was $z\sin(C)$ . My mistake. Maybe there is a mistake in my assumption that it can be expressed in that way. So I'll give the reason behind it

Waves can be expressed as $$y_{1}=A_{1}\sin(kx+\omega t+\delta_{1})$$ and $$y_{2}=A_{2}\sin(kx+\omega t+\delta_{2})$$

Let net be $y$

Therefore, $$y=y_{1}+y_{2}$$

Clearly, $$\frac{\partial y}{\partial t} = -k\frac{\partial y}{\partial x}$$

So, y can be expressed as a linear combination of x and t (leading to the conclusion that sum of two propogatory waves in the same direction fives another propogatory wave). Its probably a sine or cos function as we are adding up two sines. While we were doing superposition of waves, we assumed same amplitude for mathematical simplicity but I can vaguely remember my teacher telling me it was trivial for different amplitudes but I can't figure it out now.

Sorry if I've made a stupid mistake as I am still a beginner in waves.

Edit again: -k in partial differential equation.

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  • $\begingroup$ None can be seen on Wolframalpha: wolframalpha.com/input/?i=xSinA%2BySinB in terms of alternate representations $\endgroup$ – S.C.B. Mar 12 '16 at 14:13
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    $\begingroup$ If $x=y=1$ there seems to be a hope as there is a formula for sina+sinb $\endgroup$ – Archis Welankar Mar 12 '16 at 14:16
  • $\begingroup$ If $x^2+y^2\ne 1$ you will need complex numbers for this. $\endgroup$ – DanielWainfleet Mar 12 '16 at 14:27
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    $\begingroup$ @user254665 : What you have in mind is not at all clear to me. If you have $x\sin A + y\sin B$, you can write that as $$\sqrt{x^2+y^2}\left( \frac x {\sqrt{x^2+y^2}} \sin A + \frac y {\sqrt{x^2+y^2}} \sin B \right) = \sqrt{x^2+y^2}(u\sin A + v\sin B),$$ and then $u^2+v^2=1$. Then you could say something like $u=\cos\alpha$ and $v=\sin\alpha$. But where do you go from there? If you'd had $\sin A$ and $\cos A$ instead of $\sin A$ and $\sin B$, it would be easy, but it said $\sin A$ and $\sin B$, where $A$ and $B$ are different variables.Why would the fact that $x^2+y^2\ne1$ matter? $\qquad$ $\endgroup$ – Michael Hardy Mar 12 '16 at 14:31
  • $\begingroup$ @MichaelHardy. The Q was to write $x\sin A+y\sin B=\sin C$, not $z\sin C.$ If $0\ne x^2+y^2\ne 1$ this can be done but not with real $C$. $\endgroup$ – DanielWainfleet Mar 12 '16 at 14:41
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I'll try to answer the question you wanted to ask (as indicated in the edited question) rather than the question as originally written, whose answer you presumably already knew.

The question is, if $y_1$ and $y_2$ are the following functions of $x$ and $t$, \begin{align} y_1 &= A_1 \sin(kx + \omega t + \delta_1) \\ y_2 &= A_2 \sin(kx + \omega t + \delta_2), \end{align} how do we find a definition of $y_1+y_2$ as a function of $x$ and $t$ in the form $$ y_1 + y_2 = A_3 \sin(kx + \omega t + \delta_3)? $$


To keep the formulas a little more manageable, let $u = kx + \omega t$. Using the angle-sum formula for sine, \begin{align} \sin(kx + \omega t + \delta_1) &= \sin\delta_1 \cos u + \cos\delta_1 \sin u, \\ \sin(kx + \omega t + \delta_2) &= \sin\delta_2 \cos u + \cos\delta_2 \sin u. \end{align} Therefore \begin{align} y_1 + y_2 &= A_1(\sin\delta_1 \cos u + \cos\delta_1 \sin u) + A_2(\sin\delta_2 \cos u + \cos\delta_2 \sin u) \\ &= (A_1\sin\delta_1 + A_2\sin\delta_2)\cos u + (A_1\cos\delta_1 + A_2\cos\delta_2)\sin u. \end{align} So if $A = A_1\sin\delta_1 + A_2\sin\delta_2$ and $B = A_1\cos\delta_1 + A_2\cos\delta_2$, we're looking for $A_3$ and $\delta_3$ that will generally satisfy $$ A_3 \sin(u + \delta_3) = A \cos u + B \sin u. $$ Working from this answer to a related question, we want $A_3$ and $\delta_3$ such that $(A_3)^2 = A^2 + B^2$ and $\tan\delta_3 = \frac AB$. Recognizing that once we find a suitable $\delta_3$, we can always add or subtract any whole multiple of $2\pi$ without changing the truth of any of our equations, the solution is something of the form \begin{gather} A_3 = \pm\sqrt{A^2 + B^2}, \\ \delta_3 = \arctan \frac AB \quad\text{or}\quad \delta_3 = \pi + \arctan \frac AB. \end{gather} But since $A_3 \sin\left(u + \left(\pi + \arctan \frac AB\right)\right) = -A_3 \sin\left(u + \left(\arctan \frac AB\right)\right)$, we really have only two solutions to consider, and might as well set $A_3$ to the positive square root of $A^2 + B^2$. And then we merely need consider what happens when $u=0$, that is, check whether $A_1 \sin(\delta_1) + A_2 \sin(\delta_2)$ is positive or negative, to decide whether $\sin(\delta_3)$ must be positive or negative and therefore whether to set $\delta_3 = \arctan \frac AB$ or $\delta_3 = \pi + \arctan \frac AB$.


Alternatively, as one of my physics professors, Ken Lane, liked to say, "let's complexify" the problem. We know that in general, $$ e^{i\theta} = \cos \theta + i \sin \theta. $$ If we take the real and imaginary parts of this function, we have two sinusoidal functions $\frac\pi2$ radians out of phase. As I recall, Prof. Lane liked to take the real part of this function as the "wave", which is convenient if we want a phase-shifted cosine, but you're looking for a phase-shifted sine, so it will be more convenient to take the imaginary part. The imaginary part of $e^{i\theta}$ is $\newcommand{\Im}{\mathop{\mathrm{Im}}} \Im\left(e^{i\theta}\right) = \sin\theta$, and $$ \Im\left(A_k e^{i(kx + \omega t + \delta_k)}\right) = A_k \sin(kx + \omega t + \delta_k). $$ So we can rewrite $y_1$ and $y_2$ as the imaginary parts of \begin{align} z_1 &= A_1 e^{i(kx + \omega t + \delta_1)}, \\ z_2 &= A_2 e^{i(kx + \omega t + \delta_2)}. \end{align} Adding the two functions together, \begin{align} z_1 + z_2 &= A_1 e^{i(kx + \omega t + \delta_1)} + A_2 e^{i(kx + \omega t + \delta_2)} \\ &= \left(A_1 e^{i\delta_1} + A_2 e^{i\delta_2}\right) e^{i(kx + \omega t)} \\ &= (A_1 \cos\delta_1 + A_2 \cos\delta_2 + i (A_1 \sin\delta_1 + A_2 \sin\delta_2)) e^{i(kx + \omega t)} \end{align} Let $A = A_1 \sin\delta_1 + A_2 \sin\delta_2$ and $B = A_1 \cos\delta_1 + A_2 \cos\delta_2$; then we need merely write the complex number $B + iA$ in the form $A_3 e^{i\delta_3}$, which we do by setting $A_3 = \sqrt{A^2 + B^2}$ and either $\delta_3 = \arctan\frac AB$ (if $A$ is positive) or $\delta_3 = \pi + \arctan\frac AB$ (if $A$ is negative). We then have $$ z_1 + z_2 = \left( A_3 e^{i \delta_3} \right) e^{i(kx + \omega t)} = A_3 e^{i(kx + \omega t + \delta_3)} $$ and $$ y_1 + y_2 = \Im\left( z_1 + z_2 \right) = A_3 \sin(kx + \omega t + \delta_3). $$

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